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Question Number 123141 by ajfour last updated on 23/Nov/20

Commented by ajfour last updated on 23/Nov/20

Find side length of the congruent squares if BC=1 and AC such that two such congruent squares be possible. ∠B = 90°.

$${Find}\:{side}\:{length}\:{of}\:{the}\:{congruent} \\ $$$${squares}\:{if}\:{BC}=\mathrm{1}\:{and}\:{AC}\:{such} \\ $$$${that}\:{two}\:{such}\:{congruent}\:{squares} \\ $$$${be}\:{possible}.\:\angle{B}\:=\:\mathrm{90}°. \\ $$

Answered by mr W last updated on 23/Nov/20

∠C=θ AB=tan θ (s/((1/2)−(s/2)))=(((1/2)tan θ−s)/(s/2)) ((2s)/(1−s))=((tan θ−2s)/s) ⇒s=((tan θ)/(2+tan θ)) similarly ⇒s=((1/(tan θ))/(2+(1/(tan θ))))=(1/(2 tan θ+1)) (1/(2 tan θ+1))=((tan θ)/(2+tan θ)) ⇒tan θ=1 ⇒AB=BC

$$\angle{C}=\theta \\ $$$${AB}=\mathrm{tan}\:\theta \\ $$$$\frac{{s}}{\frac{\mathrm{1}}{\mathrm{2}}−\frac{{s}}{\mathrm{2}}}=\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\theta−{s}}{\frac{{s}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{2}{s}}{\mathrm{1}−{s}}=\frac{\mathrm{tan}\:\theta−\mathrm{2}{s}}{{s}} \\ $$$$\Rightarrow{s}=\frac{\mathrm{tan}\:\theta}{\mathrm{2}+\mathrm{tan}\:\theta} \\ $$$${similarly} \\ $$$$\Rightarrow{s}=\frac{\frac{\mathrm{1}}{\mathrm{tan}\:\theta}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{tan}\:\theta}}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{tan}\:\theta+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{tan}\:\theta+\mathrm{1}}=\frac{\mathrm{tan}\:\theta}{\mathrm{2}+\mathrm{tan}\:\theta} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\mathrm{1} \\ $$$$\Rightarrow{AB}={BC} \\ $$

Commented by ajfour last updated on 23/Nov/20

Thanks Sir, silly question.

$${Thanks}\:{Sir},\:{silly}\:{question}. \\ $$

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